Most quadratic word problems
should seem very familiar, as they are built from the linear problems
that you've done in the past.

A picture has a height
that is 4/3
its width. It is to be enlarged to have an area of 192
square inches. What will be the dimensions of the enlargement?

The height is defined
in terms of the width, so I'll pick a variable for "width",
and then create an expression for the height.

Let "w"
stand for the width of the picture. The height h
is 4/3
the width, so h
= (4/3)w. Then
the area is A
= hw = [(4/3)w][w] = (4/3)w2
= 192. I need to solve
this "area" equation for the value of the width, and then
back-solve to find the value of the height.

(4/3)w2
= 192w2
= 144w
= ± 12

Since I can't have a
negative width, I can ignore the "w
= –12" solution.
Then the width must be 12
and the height is h
= (4/3)(12) = 16.

The enlargement will
be 12
inches by 16
inches.

The product of two
consecutive negative integers is 1122.
What are the numbers?

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Remember
that consecutive integers are one unit apart, so my numbers are n
and n
+ 1. Multiplying to
get the product, I get:

The solutions are n
= –34 and n
= 33. I need a negative
value, so I'll ignore "n
= 33" and take
n
= –34. Then the other
number is n
+ 1 = (–34) + 1 = –33.

The two numbers are
–33
and –34.

Note that the second value
could have been gotten by changing the sign on the extraneous solution.
Warning: Many students get in the very bad habit of arbitrarily changing
signs to get the answers they need, but this does not always work, and
will very likely get them in trouble later on. Take the extra half a second
to find the right answer the right way.

A garden measuring
12
meters by 16
meters is to have a pedestrian pathway installed all around it, increasing
the total area to 285
square meters. What will be the width of the pathway?

The
first thing I need to do is draw a picture. Since I don't know
how wide the path will be, I'll label the width as "x".

Looking at my picture,
I see that the total width will bex
+ 12 + x = 12 + 2x,
and the total length will be x
+ 16 + x = 16 + 2x.

Then the new area
is given by:

(12 + 2x)(16
+ 2x) = 285192
+ 56x + 4x2 = 2854x2
+ 56x – 93 = 0

This quadratic is
messy enough that I won't bother with trying to use factoring to solve;
I'll just go straight to the Quadratic
Formula:

Obviously the negative
value won't work in this context, so I'll ignore it. Checking the original
exercise to verify what I'm being asked to find, I notice that I need
to have units on my answer:

The width of the
pathway will be 1.5
meters.

You have to make a
square-bottomed, unlidded box with a height of three inches and
a volume of approximately 42
cubic inches. You will be taking a piece of cardboard, cutting three-inch
squares from each corner, scoring between the corners, and folding up
the edges. What should be the dimensions of the cardboard, to the nearest
quarter inch?

When dealing with
geometric
sorts of word problems, it is usually helpful to draw a picture.
Since I'll be cutting equal-sized squares out of all of the corners,
and since the box will have a square bottom, I know I'll be starting
with a square piece of cardboard.

I don't know how
big the cardboard will be yet, so I'll label the sides as having
length "w".

Since I know I'll be cutting out
three-by-three squares to get sides that are three inches high,
I can mark that on my drawing.

The dashed lines show where I'll
be scoring the cardboard and folding up the sides.

Since I'll be losing three inches
on either end of the cardboard when I fold up the sides, the final
width of the bottom will be the original "w"
inches, less three on the one side and another three on the other
side. That is, the width of the bottom will be w
– 3 – 3 = w
– 6.

Then the volume of the box, from
the drawing, is:

(w
– 6)(w
– 6)(3) = 42(w
– 6)(w
– 6) = 14(w
– 6)2 = 14

This is the quadratic I need to solve.
I can take the square root of either side, and then add the to the right-hand
side:

Either way, I get two solutions which,
when expressed in practical decimal terms, tell me that the width of
the original cardboard is either about 2.26
inches or else about 9.74
inches.

How do I know which solution value for
the width is right? By checking each value in the original word problem.
If the cardboard is only 2.26
inches wide, then how on earth would I be able to fold up three-inch-deep
sides? But if the cardboard is 9.74
inches, then I can fold up three inches of cardboard on either side,
and still be left with 3.74
inches in the middle. Checking:

(3.74)(3.74)(3) = 41.9628

This isn't exactly 42,
but, taking round-off error into account, it's close enough that I can
trust that I have the correct value:

The cardboard should
measure 9.75
inches on a side.

In this last exercise above, you should
notice that each solution method gave the same final answer for the cardboard's
width. But the Quadratic Formula took longer and provided me with more
opportunities to make mistakes. Warning: Don't get stuck in the rut of
always using the Quadratic Formula!